A point $P$ is randomly placed in the interior of the right triangle below.  What is the probability that the area of triangle $PBC$ is less than half of the area of triangle $ABC$?  Express your answer as a common fraction. [asy]
size(7cm);
defaultpen(linewidth(0.7));
pair A=(0,5), B=(8,0), C=(0,0), P=(1.5,1.7);
draw(A--B--C--cycle);
draw(C--P--B);
label("$A$",A,NW);
label("$B$",B,E);
label("$C$",C,SW);
label("$P$",P,N);
draw((0,0.4)--(0.4,0.4)--(0.4,0));[/asy]
Explanation: Let $h$ be the distance from $P$ to $CB$.  The area of triangle $ABC$ is $\frac{1}{2}(AC)(CB)$.  The area of triangle $PBC$ is $\frac{1}{2}(h)(CB)$.  Therefore, the area of triangle $PBC$ is less than one-half of the area of triangle $ABC$ if $h<AC/2$.  This happens if $P$ falls below the dashed line whose endpoints are the midpoints $D$ and $E$ of $\overline{AC}$ and $\overline{AB}$.  Triangle $ADE$ is similar to triangle $ACB$, so the ratio of the area of triangle $ADE$ to the area of triangle $ACB$ is $\left(\frac{AD}{AC}\right)^2=\frac{1}{4}$.  Therefore, the ratio of the area of trapezoid $DEBC$ to the area of triangle $ABC$ is $1-\dfrac{1}{4}=\boxed{\frac{3}{4}}$.

[asy]
size(7cm);
defaultpen(linewidth(0.7));
pair A=(0,5), B=(8,0), C=(0,0), P=(1.5,1.7);
draw(A--B--C--cycle);
draw((A+C)/2--(A+B)/2,dashed);
dot(P);
label("$A$",A,NW);
label("$B$",B,E);
label("$C$",C,SW);
label("$P$",P,E);
label("$D$",(A+C)/2,W);
label("$E$",(A+B)/2,NE);
draw((0,0.4)--(0.4,0.4)--(0.4,0));[/asy]